Prove that $\mathbb Q(\sqrt 2, i) = \mathbb Q(\sqrt 2 + i)$ I probably don't understand the definition of each properly. What I am thinking is that 
$\mathbb Q(\sqrt 2, i)$ has elements of form $a_1(\sqrt2) ^1 + a_2(\sqrt2) ^2 + ... + a_n(\sqrt2) ^n + a_1i^1 + a_2i^2 + ... + a_ni^n$
$\mathbb Q(\sqrt 2 +i)$ has elements of form $a_1(\sqrt 2 + i)^1 + a_2(\sqrt 2 + i)^2 + ... + a_n(\sqrt 2 + i)^n$
I'm having trouble jumping from one to the other, as by this definition $\mathbb Q(\sqrt 2 +i)$ has elements of form $\sqrt 2 ^ni^m$ and it seems as though $\mathbb Q(\sqrt 2, i)$ doesn't. Can anyone help?
 A: You are mistaken in what $\mathbb Q(\sqrt 2,i)$ is. It not only contains linear combinations of all powers of $\sqrt 2$ and of $i$, but also of elements of the form $\sqrt 2^mi^n$.
A: Hints:
(i)  Clearly $\;\Bbb Q(\sqrt2+i)\le\Bbb Q(\sqrt2,i)\;$ (why?)
(ii)  $\;\dim_{\Bbb Q}\Bbb Q(\sqrt2,i)=4\;$ (why? You may want to prove that $\;x^2+1\in\Bbb Q(\sqrt2)[x]\;$ is irreducible...)
$$\text{(iii)}\;\;x=\sqrt2+i\implies x^2-2\sqrt2\,x+2=-1\implies x^2+3=2\sqrt2\,x\implies$$
$$\implies x^4+6x^2+9=8x^2\implies x^4-2x^2+9=0$$
(iv) Thus $\;\dim_{\Bbb Q}\Bbb Q(\sqrt2+i)=4\;$ (why?) Show the above quartic is irreducible over the rationals).
Now solve you problem...
A: $\mathbb Q(\sqrt 2, i)$ is the smallest subfield of $\mathbb C$ which contains the elements $\sqrt 2$ and $i$. So, in particular, it contains the product $\sqrt 2 \cdot i$ of these two elements. Similarly, $\mathbb Q(\sqrt 2 + i)$ is the smallest subfield of $\mathbb C$ containing the element $\sqrt 2 +i$. In particular then, it contains the element $(\sqrt 2 + i)^2=1 + 2\sqrt 2\cdot i$. Since $1$ and $2$ are certainly in $\mathbb Q(\sqrt 2 + i)$, it follows that $\sqrt 2 \cdot i\in \mathbb Q (\sqrt 2 + i)$. But then it follows that $(\sqrt 2 + i)\cdot \sqrt 2 \cdot i=2i - \sqrt 2$ is in $\mathbb Q (\sqrt 2 + i)$. And then $2i-\sqrt 2 + \sqrt 2 + i=3i$ is in $\mathbb Q(\sqrt 2 + i)$, and finally also $i$ is there. You may now easily conclude that $\sqrt 2 \in \mathbb Q (\sqrt 2 + i)$. Now, this shows that $\mathbb Q (\sqrt 2 , i)\subseteq \mathbb Q (\sqrt 2 + i)$. The other inclusion is obtained similarly (but much more quickly and easily). 
A: By definition, $\mathbb Q(a_i,i\in I)$ is the smallest field containing $\mathbb Q$ and $a_i$ for all $i$.
So, $\mathbb Q(\sqrt{2},i)$ is a field containing $\mathbb Q$, $\sqrt{2}$ and $i$ so it also contains $\sqrt{2}+i$. Hence you get a first inclusion : $$\mathbb Q(\sqrt{2}+i) \subset \mathbb Q(\sqrt{2},i).$$
On the other hand, if $x$ denotes $\sqrt{2}+i$, you can check that $\dfrac{1}{6}(x^3+x)=i$ so $i\in \mathbb Q(\sqrt{2}+i)$ and consequently $\sqrt{2} \in \mathbb Q(\sqrt{2}+i)$.
So $$\mathbb Q(\sqrt{2},i) \subset \mathbb Q(\sqrt{2}+i).$$
A: Try playing with the elements. For example,  $$(\sqrt 2+i)^3=2\sqrt 2+6i-3\sqrt                                  2-i=-\sqrt2+5i$$
Thus, it follows that $(\sqrt 2+i)-(\sqrt 2+5i)=-4i\in\mathbb Q(\sqrt2+i)$. It should be clear that this implies that $i$ and hence $\sqrt2$ belong to $\mathbb Q(\sqrt2+i)$.
